Question

The probability of randomly selecting a red marble from a bag, replacing it, and selecting a red marble again is 1/25.
What is the probability of selecting a red marble on the first try?

Answer 1

\[\frac{1}{25}\]

Answer 2

I will become your fan and give you a medal for the right answer

Answer 3

reason: same number of marbles in the bag the second time as the first. same number of red marbles the second time as well. so nothing has changed and the probabilities are the same.

Answer 4

oh thanks!

Answer 5

oh wait. i need to read the question more carefully. hold on.

Answer 6

hold on i think i was mistaken. sorry.

Answer 7

i misread.
let me try again and this time go slow, and maybe with an explanation.

Answer 8

ok

Answer 9

put A = the event you select a red marble on the first try. we want
\[P(A)\]
B = event you select a red marble on second try.

Answer 10

\[A\cap B\] is the even that both marbles selected were red, and you are given that \[P(A\cap B)=\frac{1}{25}\] because you are given that the probability that both marbles selected were red

Answer 11

in general
\[P(A\cap B) = P(A)\times P(B|A)\]
but since you have replaced the red marble there are the same number of red marble in the bag and the same number of marbles, so \[P(B)=P(A)\]

Answer 12

Shouldn't it be 1/5. 1/5 * 1/5 = 1/25
th

Answer 13

therefore you have
\[\frac{1}{25}=P(A)\times P(A)\] so \[P(A)=\frac{1}{5}\]

Answer 14

That is the probability of selecting it once into the same probability of it being picked again

Answer 15

yes you are right isan_meswani. my first answer was a mistake. i read the problem incorrectly.

Answer 16

1/5?
not1/25?

Answer 17

yes, 1/5

Answer 18

sorry i messed you up on first try.

Answer 19

Suppose you know there are originally 2 red marbles in the bag. How many marbles are there in all?

Answer 20

Now this is simple, isn't it?
its 10

Answer 21

really?

Answer 22

yes that was an easy one.
\[\frac{1}{5}=\frac{2}{x}\]

Answer 23

oh thanks!